newton raphson & jacobi
DESCRIPTION
psaTRANSCRIPT
Power Systems I
The Power Flow Solution
l Most common and important tool in power systemanalysisu also known as the “Load Flow” solutionu used for planning and controlling a systemu assumptions: balanced condition and single phase analysis
l Problem:u determine the voltage magnitude and phase angle at each busu determine the active and reactive power flow in each lineu each bus has four state variables:
n voltage magnituden voltage phase anglen real power injectionn reactive power injection
Power Systems I
The Power Flow Solution
u Each bus has two of the four state variables defined or givenl Types of buses:
u Slack bus (swing bus)n voltage magnitude and angle are specified, reference busn solution: active and reactive power injections
u Regulated bus (generator bus, P-V bus)n models generation-station busesn real power and voltage magnitude are specifiedn solution: reactive power injection and voltage angle
u Load bus (P-Q bus)n models load-center busesn active and reactive powers are specified (negative values for loads)n solution: voltage magnitude and angle
Power Systems I
Newton-Raphson PF Solution
l Quadratic convergenceu mathematically superior to Guass-Seidel method
l More efficient for large networksu number of iterations required for solution is independent of
system sizel The Newton-Raphson equations are cast in natural power
system formu solving for voltage magnitude and angle, given real and reactive
power injections
Power Systems I
Newton-Raphson Method
l A method of successive approximation using Taylor’sexpansionu Consider the function: f(x) = c, where x is unknown
u Let x[0] be an initial estimate, then ∆x[0] is a small deviation fromthe correct solution
u Expand the left-hand side into a Taylor’s series about x[0] yeilds
( ) cxxf =∆+ ]0[]0[
( ) ( ) cxdx
fdxdxdfxf =+∆
+∆
+ L
2]0[2
2
21]0[]0[
Power Systems I
Newton-Raphson Method
u Assuming the error, ∆x[0], is small, the higher-order terms areneglected, resulting in
u where
u rearranging the equations
( ) ]0[]0[]0[]0[ xdxdfccx
dxdfxf ∆
≈∆⇒≈∆
+
( )]0[]0[ xfcc −=∆
]0[]0[]1[
]0[]0[
xxx
dxdfcx
∆+=
∆=∆
Power Systems I
Examplel Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0
Power Systems I
Newton-Raphson Method
0 1 2 3 4 5 6-10
0
10
20
30
40
50
x
f(x) = x3-6x2+9x-4
Power Systems I
( )∑
∑∑
=
==
+∠−∠=−
=−
+∠==
n
jjijjijiii
iiii
n
jjijjij
n
jjiji
VYVQjP
IVQjP
VYVYI
1
*
11
δθδ
δθ
Power Flow Equations
l KCL for current injection
l Real and reactive power injection
l Substituting for Ii yields:
Power Systems I
Power Flow Equations
( )
( )∑
∑
=
=
+−−=
+−=
n
jjiijijjii
n
jjiijijjii
YVVQ
YVVP
1
1
sin
cos
δδθ
δδθ
l Divide into real and reactive parts
Power Systems I
Newton-Raphson Formation
( )
( )
( ) ( )( )
=
=
=
+−−=
+−=
∑
∑
=
=
][
][][
][
][][
1
][][][][][
1
][][][][][
sin
cos
kinj
kinjk
k
kk
schinj
schinj
n
j
kj
kiijij
kj
ki
ki
n
j
kj
kiijij
kj
ki
ki
xQxP
xfV
xQP
c
YVVQ
YVVP
δ
δδθ
δδθ
l Cast power equations into iterative form
l Matrix function formation of the system of equations
Power Systems I
Newton-Raphson Formation
( )
( )( )
( )dx
xdf
dxxdf
xfcxx
xxxfc
k
k
kkk
solutionsolution
][
][
][][]1[
]0[ of estimate initial
−+=
==
+
l General formation of the equation to find a solution
l The iterative equation
l The Jacobian - the first derivative of a set of functions
a matrix of all combinatorial pairs
Power Systems I
The Jacobian Matrix
( )
∆
∆∆
∆
=
∆
∆∆
∆
∆∆
∂∂
∂∂
∂∂
∂∂
=
∆∆
⇒
−
−
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
−
−
−
−−
−
−−
−−
−
−−
−
−−
−−
mn
n
VQ
VQQQ
VQ
VQQQ
VP
VPPP
VP
VPPP
mn
n
V
V
Q
QP
P
VV
QQV
PP
QP
dxxdf
mn
mnmn
n
mnmn
mnn
mn
nn
n
nn
mnn
M
M
LLMOMMOM
LL
LLMOMMOM
LL
M
M
1
1
1
1
1
1
111
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
δ
δ
δ
δ
δ
δδ
δδ
δδ
δδ
Power Systems I
Jacobian Terms
( )
( )
( )
( ) jiYVVP
YVYVVP
jiYVVP
YVVP
jiijijij
i
ijjiijijjiiiii
i
i
jiijijjij
i
ijjiijijji
i
i
≠+−=∂∂
+−+=∂∂
≠+−−=∂∂
+−=∂∂
∑
∑
≠
≠
δδθ
δδθθ
δδθδ
δδθδ
cos
coscos2
sin
sin
l Real power w.r.t. the voltage angle
l Real power w.r.t. the voltage magnitude
Power Systems I
l Reactive power w.r.t. the voltage angle
l Reactive power w.r.t. the voltage magnitude
Jacobian Terms
( )
( )
( )
( ) jiYVVQ
YVYVVQ
jiYVVQ
YVVQ
jiijijij
i
ijjiijijjiiiii
i
i
jiijijjij
i
ijjiijijji
i
i
≠+−−=∂∂
+−+−=∂∂
≠+−−=∂∂
+−=∂∂
∑
∑
≠
≠
δδθ
δδθθ
δδθδ
δδθδ
sin
sinsin2
cos
cos
Power Systems I
Iteration process
l Power mismatch or power residualsu difference in schedule to calculated power
l New estimates for the voltages
][][]1[
][][]1[
][][
][][
ki
ki
ki
ki
ki
ki
ki
schi
ki
ki
schi
ki
VVV
QQQ
PPP
∆+=∆+=
−=∆−=∆
+
+ δδδ
Power Systems I
Bus Type and the Jacobian Formation
l Slack Bus / Swing Busu one generator bus must be selected and defined as the voltage
and angular referencen The voltage and angle are known for this busn The angle is arbitrarily selected as zero degreesn bus is not included in the Jacobian matrix formation
l Generator Busn have known terminal voltage and real (actual) power injectionn the bus voltage angle and reactive power injection are computedn bus is included in the real power parts of the Jacobian matrix
l Load Busn have known real and reactive power injectionsn bus is fully included in the Jacobian matrix
Power Systems I
Newton-Raphson Steps
1. Set flat startu For load buses, set voltages equal to the slack bus or 1.0∠0°u For generator buses, set the angles equal the slack bus or 0°
2. Calculate power mismatchu For load buses, calculate P and Q injections using the known and
estimated system voltagesu For generator buses, calculate P injectionsu Obtain the power mismatches, ∆P and ∆Q
3. Form the Jacobian matrixu Use the various equations for the partial derivatives w.r.t. the
voltage angles and magnitudes
Power Systems I
Newton-Raphson Steps
4. Find the matrix solution (choose a or b)u a. inverse the Jacobian matrix and multiply by the mismatch
poweru b. perform gaussian elimination on the Jacobian matrix with the b
vector equal to the mismatch powercompute ∆δ and ∆V
5. Find new estimates for the voltage magnitude and angle6. Repeat the process until the mismatch (residuals) are
less than the specified accuracy
ε
ε
≤∆
≤∆][
][
ki
ki
Q
P
Power Systems I
Line Flows and Losses
l After solving for bus voltages and angles, power flowsand losses on the network branches are calculatedu Transmission lines and transformers are network branchesu The direction of positive current flow are defined as follows for a
branch element (demonstrated on a medium length line)u Power flow is defined for each end of the branch
n Example: the power leaving bus i and flowing to bus j
VjVi
yj0yi0
yijBus i Bus j
Iij IjiIL
Ij0Ii0
Power Systems I
Line Flows and Losses
l current and power flows:
l power loss:
VjVi
yj0yi0
yijBus i Bus j
Iij IjiIL
Ij0Ii0
( )( ) ***
02*
00
jijiiijiijiij
iijiijiLij
VyVyyVIVS
VyVVyIIIji
−+==
+−=+=→
( )( ) ***
02*
00
iijjjijjjijji
jjijijjLji
VyVyyVIVS
VyVVyIIIij
−+==
+−=+−=→
jiijijLoss SSS +=
Power Systems I
Example
j0.04
3
1
2
138.6 MW 45.2 MVAR
256.6 MW 110.2 MVAR
Slack BusV1 = 1.05∠0°
j0.02j0.025
l Using N-R method, find thephasor voltages at buses 2and 3
l Find the slack bus realand reactive power
l Calculate line flowsand line lossesu 100 MVA base